MATLAB Linear Programming. Greg Reese, Ph.D Research Computing Support Group Academic Technology Services Miami University

Transcription

1 MATLAB Linear Programming Greg Reese, Ph.D Research Computing Support Group Academic Technology Services Miami University

2 MATLAB Linear Programming Greg Reese. All rights reserved

3 Optimization Optimization - finding value of a parameter that maimizes or minimizes a function with that parameter Talking about mathematical optimization, not optimization of computer code! "function" is mathematical function, not MATLAB language function 3

4 Optimization Optimization Can have multiple parameters Can have multiple functions Parameters can appear linearly or nonlinearly

5 Linear programming Most often used kind of optimization Tremendous number of practical applications "Programming" means determining feasible programs (plans, schedules, allocations) that are optimal with respect to a certain criterion and that obey certain constraints 5

6 A feasible program is a solution to a linear programming problem and that satisfies certain constraints In linear programming Constraints are linear inequalities Criterion is a linear epression Epression called the objective function In practice, objective function is often the cost of or profit from some activity 6

7 Many important problems in economics and management can be solved by linear programming Some problems are so common that they're given special names 7

8 DIET PROBLEM You are given a group of foods, their nutritional values and costs. You know how much nutrition a person needs. What combination of foods can you serve that meets the nutritional needs of a person but costs the least? 8

9 BLENDING PROBLEM Closely relate to diet problem Given quantities and qualities of available oils, what is cheapest way to blend them into needed assortment of fuels? 9

10 TRANSPORTATION PROBLEM You are given a group of ports or supply centers of a certain commodity and another group of destinations or markets to which commodity must be shipped. You know how much commodity at each port, how much each market must receive, cost to ship between any port and market. How much should you ship from each port to each market so as to minimize the total shipping cost? 0

11 WAREHOUSE PROBLEM You are given a warehouse of known capacity and initial stock size. Know purchase and selling price of stock. Interested in transactions over a certain time, e.g., year. Divide time into smaller periods, e.g., months. How much should you buy and sell each period to maimize your profit, subject to restrictions that. Amount of stock at any time can't eceed warehouse capacity. You can't sell more stock than you have

12 Linear programming Mathematical formulation The variables,,... n satisfy the inequalities and 0, 0,... n 0. Find the set of values of,,... n that minimizes (maimizes) Note that a pq and f i are known m n mn m m n n n n b a a a b a a a b a a a n f n f f

13 3 Linear programming Mathematical matri formulation Find the value of that minimizes (maimizes) f T given that 0 and A b, where n n m mn m m n f f f f b b b b a a a a a a a a a A and,,,

14 General procedure Linear programming. Restate problem in terms of equations and inequalities. Rewrite in matri and vector notation 3. Call MATLAB function linprog to solve

15 Eample - diet problem My son's diet comes from the four basic food groups - chocolate dessert, ice cream, soda, and cheesecake. He checks in a store and finds one of each kind of food, namely, a brownie, chocolate ice cream, Pepsi, and one slice of pineapple cheesecake. Each day he needs at least 500 calories, 6 oz of chocolate, 0 oz of sugar, and 8 oz of fat. Using the table on the net slide that gives the cost and nutrition of each item, figure out how much he should buy and eat of each of the four items he found in the store so that he gets enough nutrition but spends as little (of my money...) as possible. 5

16 Eample - diet problem Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice 6

17 Eample - diet problem What are unknowns? Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice = number of brownies to eat each day = number of scoops of chocolate ice cream to eat each day 3 = number of bottles of Coke to drink each day = number of pineapple cheesecake slices to eat each day In linear programming "unknowns" are called decision variables 7

18 Eample - diet problem Objective is to minimize cost of food. Total daily cost is Cost = (Cost of brownies) + (Cost of ice cream) + (Cost of Coke) + (Cost of cheesecake) Cost of brownies = (Cost/brownie) (brownies/day) =.5 Cost of ice cream = Cost of Coke =.5 3 Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice Cost of cheesecake = 8

19 Eample - diet problem Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice Therefore, need to minimize

20 Eample - diet problem Constraint - calorie intake at least 500 Calories from brownies = (calories/brownie)(brownies/day) = 00 Calories from ice cream = 00 Calories from Coke = 50 3 Calories from cheesecake = 500 So constraint is Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice

21 Eample - diet problem Constraint - chocolate intake at least 6 oz Chocolate from brownies = (Chocolate/brownie)(brownies/day) = 3 Chocolate from ice cream = Chocolate from Coke = 0 3 = 0 Chocolate from cheesecake = 0 = 0 So constraint is Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice 3 6

22 Eample - diet problem Constraint 3 - sugar intake at least 0 oz Sugar from brownies = (sugar/brownie)(brownies/day) = Sugar from ice cream = Sugar from Coke = 3 Sugar from cheesecake = So constraint 3 is Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice 3 0

23 Eample - diet problem Constraint - fat intake at least 8 oz Fat from brownies = (fat/brownie)(brownies/day) = Fat from ice cream = Fat from Coke = 3 Fat from cheesecake = 5 So constraint is Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice

24 Eample - diet problem Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice Finally, we assume that the amounts eaten are non-negative, i.e., we ignore throwing up. This means that we have 0, 0, 3 0, and 0

25 Eample - diet problem Putting it all together, we have to minimize subject to the constraints and Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice

26 Eample - diet problem Food Calories Chocolate Sugar (ounces) Fat (ounces) Cost (serving) Brownie 00 3 $.50 / brownie Chocolate ice cream 00 $.00 / scoop Coke 50 0 $.50 / bottle Pineapple cheesecake $.00 / slice In matri notation, want to minimize f T subject to A b and 0 where A , 5 b 500 6, 0 8 3, and f.5.5 6

27 MATLAB solves linear programming problem minimize f T A b such that Aeq beq lb ub where, b, beq, lb, and ub are vectors and A and Aeq are matrices. Can use one or more of the constraints "lb" means "lower bound", "ub" means "upper bound" Often have lb = 0 and ub =, i.e., no upper bound 7

28 MATLAB linear programming solver is linprog(), which you can call various ways: = linprog(f,a,b) = linprog(f,a,b,aeq,beq) = linprog(f,a,b,aeq,beq,lb,ub) = linprog(f,a,b,aeq,beq,lb,ub,0) = linprog(f,a,b,aeq,beq,lb,ub,0,options) = linprog(problem) [,fval] = linprog(...) [,fval,eitflag] = linprog(...) [,fval,eitflag,output] = linprog(...) [,fval,eitflag,output,lambda] = linprog(...) 8

29 9 Linear programming Eample - diet problem Us: MATLAB: Note two differences:.5.5 and,, , f b A b A f T 0 and subject to minimize such that minimize ub lb beq Aeq b A f T

30 Eample - diet problem ISSUE - We have A b but need A b One way to handle is to note that if A b then -A -b, so can have MATLAB use constraint (-A) (-b) ISSUE - We have 0 but MATLAB wants lb ub. Handle by omitting ub in call of linprog(). If omitted, MATLAB assumes no upper bound 30

31 Eample - diet problem = linprog(f,a,b,aeq,beq,lb,ub) We'll actually call = linprog(f,a,b,aeq,beq,lb) If don't have equality constraints, pass [] for Aeq and beq 3

32 Eample - diet problem Follow along now >> A = -[ ; 3 0 0; ;... 5 ]; >> b = -[ ]'; >> f = [.5.5 ]'; >> lb = [ ]'; >> = linprog( f, A, b, [], [], lb ) Optimization terminated. = % brownies % chocolate ice cream.0000 % Coke % cheesecake 3

33 Eample - diet problem Optimal solution is = [ ] T. In words, my son should eat 3 scoops of ice cream and drink Coke each day. 33

34 Eample - diet problem A constraint is binding if both sides of the constraint inequality are equal when the optimal solution is substituted. For = [ ] T the set becomes 6 6, so the chocolate and sugar constraints are binding. The other two are nonbinding

35 Eample - diet problem How many calories, and how much chocolate, sugar and fat will he get each day? >> -A* ans = % calories % chocolate % sugar % fat How much money will this cost? >> f'* ans =.5000 % dollars 35

36 Eample - diet problem Because it's common to want to know the value of the objective function at the optimum, linprog() can return that to you [ fval] = linprog(f,a,b,aeq,beq,lb,ub) where fval = f T >> [ fval] = linprog( f, A, b, [], [], lb ) = fval =

37 Special kinds of solutions Usually a linear programming problem has a unique (single) optimal solution. However, there can also be:. No feasible solutions. An unbounded solution. There are solutions that make the objective function arbitrarily large (ma problem) or arbitrarily small (min problem) 3. An infinite number of optimal solutions. The technique of goal programming is often used to choose among alternative optimal solutions. (Won't consider this case more) 37

38 Can tell about the solution MATLAB finds by using third output variable: [ fval eitflag] =... linprog(f,a,b,aeq,beq,lb,ub) eitflag - integer identifying the reason the algorithm terminated. Values are Function converged to a solution. 0 Number of iterations eceeded options. - No feasible point was found. -3 Problem is unbounded. - NaN value was encountered during eecution of the algorithm. -5 Both primal and dual problems are infeasible. -7 Search direction became too small. No further progress could be made. 38

39 Try It Linear programming Solve the following problem and display the optimal solution, the value of the objective value there, and the eit flag from linprog() Maimize z = - subject to,

40 0 Linear programming Try It First multiply second equation by - to get Then, with objective function z = - rewrite in matri form: 0, and,, 6, lb f b A

41 Linear programming Try It >> A = [ -; - - ]; >> b = [ -6 ]'; >> f = [ - ]'; >> lb = [ 0 0 ]'; 0 0 and,, 6, lb f b A

42 Try It IMPORTANT - linprog() tries to minimize the objective function. If you want to maimize the objective function, pass -f and use -fval as the maimum value of the objective function

43 Try It >> [ fval eitflag] = linprog( -f, A, b, [],[], lb ) Eiting: One or more of the residuals, duality gap, or total relative error has grown times greater than its minimum value so far: the dual appears to be infeasible (and the primal unbounded). (The primal residual < TolFun=.00e-008.) =.0e+06 * fval = -.69e+06 (-fval =.69e+06!!!) eitflag = -3 (Problem is unbounded) 3

44 Try It A farmer has 0 acres to plant in wheat and rye. He has to plant at least 7 acres. However, he has only $00 to spend and each acre of wheat costs $00 to plant and each acre of rye costs $00 to plant. Moreover, the farmer has to get the planting done in hours and it takes an hour to plant an acre of wheat and hours to plant an acre of rye. If the profit is $500 per acre of wheat and $300 per acre of rye how many acres of each should be planted to maimize profits?

45 Try It Decision variables is number of acres of wheat to plant y is number of acres of rye to plant Constraints "has 0 acres to plant in wheat and rye" In math this is y 0 " has to plant at least 7 acres" In math this is y 7 5

46 Try It Constraints Linear programming "he has only $00 to spend and each acre of wheat costs $00 to plant and each acre of rye costs $00 to plant" 00 00y In math this is 00 6

47 Try It Constraints Linear programming "the farmer has to get the planting done in hours and it takes an hour to plant an acre of wheat and hours to plant an acre of rye " y In math this is 7

48 Try It Objective function Linear programming "... the profit is $500 per acre of wheat and $300 per acre of rye" z In math this is y 8

49 Try It Put it together Constraints: y y 00 00y y y Objective function: z y 9

50 Try It Rename to and y to Change + y 7 to - - y -7 and then to z

51 5 Linear programming Try It Write in matri form Maimize Maimize z 0 0 and ,, , lb f b A f z T

52 Try It Find solution that maimizes profit. Display both >> A = [ ; - -; 00 00; ]; >> b = [ ]'; >> f = [ ]'; >> lb = [ 0 0 ]'; >> [ fval] = linprog( -f, A, b, [], [], lb ); >> ' A 00, 00 ans = >> maprofit = -fval 0 7 b, 00 maprofit = 3.000e+003, f and lb z f T 5

53 Try It - blending problem Alloy Miture Optimization (minimize epenses) There are four metals with the following properties: Metal Density %Carbon %Phosphor Price ($/kg) A B C D We want to make an alloy with properties in the following range: Range Density %Carbon %Phosphor Minimum Maimum What miture of metals should we use to minimize the cost of the alloy? 53

54 Try It - blending problem Decision variables is fraction of total alloy that is metal A is fraction of total alloy that is metal B 3 is fraction of total alloy that is metal C is fraction of total alloy that is metal D 5

55 Metal Density %Carbon %Phosphor Price ($/kg) A B C Range Density %Carbon %Phosphor Minimum Maimum Try It - blending problem Density constraints Alloy density must be at least 5950 In math this is Alloy density must be at most 6050 In math this is D

56 Try It - blending problem Carbon constraints Carbon concentration must be at least 0. In math this is Carbon concentration must be at most 0.3 In math this is Metal Density %Carbon %Phosphor Price ($/kg) Range A Density %Carbon %Phosphor B Minimum C Maimum D

57 Metal Density %Carbon %Phosphor Price ($/kg) A B C Range Density %Carbon %Phosphor Minimum Maimum Try It - blending problem Phosphor constraints Phosphor concentration must be at least 0. In math this is Phosphor concentration must be at most 0.3 In math this is D

58 Try It - blending problem Constraints Since only the four metals will make up the alloy, the sum of the fractional amounts must be one: 3 Fractional parts must be non-negative: 0 for i,,3, i (Each part must also be, but that's handled by first equation.) 58

59 Metal Density %Carbon %Phosphor Price ($/kg) A B C D Try It - blending problem Objective function Cost per kg z

60 Try It - blending problem Put it together Constraints: (Convert to ) i , i,,3, Objective function: z

61 6 Linear programming Try It - blending problem Write in matri form Minimize and, b, A, , , EQ T EQ 3 lb f b A f z T,,3, 0, i i

62 Linear programming A f, A Try It - blending problem EQ T, b EQ , , and lb b, >> A = [ ; ; ; ; ; ]; >> b = [ ]'; >> f = [.5.5 ]'; >> Aeq = [ ]; >> beq = ; >> lb = [ ]'; 6

63 Try It - blending problem >> [ fval] = linprog( f, A, b, Aeq, beq, lb ) Optimization terminated. = <- Metal A 0.85 <- Metal B <- Metal C 0.07 <- Metal D fval =.88 <- Profit in $/kg 63

64 MATLAB Linear Programming Questions? 6

65 The End 65